Steel billets of 35 mm diameter and 500 mm length are to be heated to 1200°C by direct resistance heating.

Heating time (sec) 30

Density (g/cm³) 7.6

Mean thermal conductivity (λ.w/m °C) 35.7

Mean specific heat (kJ/kg °C) 0.576

Mean resistivity (σ ⋅ ohm ⋅ cm) 61 × 10^{-6}

Ambient temperature (°C) 30

Magnetic permeability (μ_r) 100

Permeability of air (μ_o) 4\pi × 10^{-7}

Now determine:

1. Power required to heat the billet from 30 to 1200°C.

2. Power lost by radiation and total power required.

3. Current required from a 100 V DC or AC supply.

4. Effective resistance for DC and AC heating.

Step-by-Step

Learn more on how do we answer questions.

Volume of billet

= =\frac{\pi \times 3.5^2 \times 50}{4}=480 \ \mathrm{~cm}^3

Surface area (curved surface)

= π × 3.5 × 50

= 550 cm²

= 0.055 m²

Weight

= 480 × 7.6

= 3648 gm

= 3.648 kg

Resistance

\begin{aligned} \sigma \frac{\ell}{A} & =\frac{61 \times 10^{-6} \times 50}{38.48} \\ & =80 \times 10^{-6} \ \Omega \end{aligned}

Heat required to raise the temperature from 30 to 1200°C = mass × sp. heat × temp. difference

= 3.648 × 10³ × 0.576 × (1200 − 30)

= 2.458 × 106 J

If heating is done in 30 sec

Power = \frac{2.458 \times 10^6}{30}

= 85 kW

Heat radiated

Assume a mean surface temperature 800°C and an emissivity 0.8

Total power

= Heating load + radiated loss

= 85 + 12

= 97 kW~100 kW

**Note : **There may be additional losses in grips.

If a 100 V DC power supply is available

Current required = \frac{100 \times 10^3}{1000}

= 1000 A

The radiation loss is very high. It can be reduced by keeping the billet in an enclosure or heating for a shorter duration.

Note that the calculated radiation loss is at 1200°C surface temperature which is attained at the end of the heating time.

If the power available is AC at 50 Hz, the effective resistance will change from R_{DC} to R_{AC}.

R_{DC} = 80 × 10^{-6} Ω (as calculated earlier)

\rho ⋅ (30–760°C) = 57× 10^{-6} ohm.cm

\rho ⋅ (760–1200°C) = 120 ×10^{-6} ohm.cm

\mu_o=4 \pi \times 10^{-7} \quad \mu_r=100

We can get R_{AC} by using the formula

R_{AC} = \rho \frac{\ell}{A}

where a is the area of annulus through which most of the current flows. This will be circular ring of outer diameter 3.5 cm and inner diameter = 3.5 − 2δ, where δ is the depth of penetration.

For the temperature range 30–760°C

= 0.53 cm or 5.3 mm

Inner diameter = 3.5-2(2 \times \delta)

= 3.5 – 2 (2× 0.53)

= 1.38 cm

Area = \frac{\pi\left(3.5^2-1.38^2\right)}{4}

= 8.125 cm²

\begin{aligned}R_{A C} & =\frac{57 \times 10^{-6} \times 50}{8.125} \\& =350 \times 10^{-6}\end{aligned}or = 35 × 10^{-5} ohm (30-760°C).

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